Answer:
2 Atm; 2.016 g
Explanation:
Changing the volume without changing the temperature or mass only changes the pressure. Volume and pressure are inversely proportional so halving the volume will double the pressure.
P = 1 Atm, T = 0 °C are "standard" temperature and pressure (STP). The volume of 1 mole of gas is 22.4 L under these conditions. That means the amount of hydrogen gas in the cylinder is 1 mole, so has a mass of 2.016 g.
After the volume reduction, the pressure is 2 Atm, and the mass remains 2.016 g.
Answer:
B 144.0 s is the best answer of this question
1st find the pOH = 14 - 4.75 = 9.25
then do 10^-9.25 = 5.62x10^-10 OH- concentration
True. Ions can have a positive or negative charge.
Answer:
Explanation:
The air 9% mole% methane have an average molecular weight of:
9%×16,04g/mol + 91%×29g/mol = 27,8g/mol
And a flow of 700000g/h÷27,8g/mol = 25180 mol/h
In the reactor where methane solution and air are mixed:
In = Out
Air balance:
91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)
Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>
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The air in the product gas is
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>
43058 mol air×29g/mol <em>1249 kg air</em>
Percent of oxygen is:
=<em>0,231 kg O₂/ kg air</em>
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I hope it helps!